I want to show the group cohomology $H^{n>1}\left(F,\,M\right)$ vanishes whenever $F$ is free.

Question

I want to show the group cohomology $H^{n>1}\left(F,\,M\right)$ vanishes whenever $F$ is free.

I tried to show $\text{pdim}_{\mathbb{Z}\left[F\right]}\left(\mathbb{Z}\right)\le 1$, but we know projective resolution $\cdots\rightarrow\mathbb{Z}\left[F^{n+1}\right]\rightarrow \mathbb{Z}\left[F^{n}\right]\rightarrow\cdots$ and $\mathbb{Z}\left[F^{3}\right]\ne 0$. It is contradict to what I want to show.

I will wait your help.

Answer 1

Let $F=F(X)$ be free over the set $X$. A free resolution of $\mathbb{Z}$ over $\mathbb{Z}[F]$ is defined by $$\cdots \to 0\to 0 \to \bigoplus_{x \in X}\mathbb{Z}[F] \xrightarrow{f} \mathbb{Z}[F] \to \mathbb{Z} \to 0$$ where $$f: (a_x)_{x \in X} \mapsto \sum_x a_x(x-1)$$ In particular, the projective dimension of $F$ is one.

A proof for this resolution is in Rotman: An Introduction to homological algebra, Prop. 9.54 and Cor. 9.55.

Answer 2

Here's a topological proof. Recall that

$$H^n(G, M) \cong H^n_{sing}(K(G, 1); M)$$

where $K(G, 1)$ is the space with with fundamental group isomorphic to $G$ and universal cover being contractible and $H^n_{sing}$ denotes the singular cohomology groups.

Let $G = F$ be a free group. Then $K(F, 1)$ is simply wedge of $|F|$-many circles, since universal cover is just a tree, which is contractible. By Mayer-Vietoris,

$$H^n_{sing}(\bigvee_{|F|}S^1; M) \cong \bigoplus_{|F|} H^n_{sing}(S^1, M) \cong 0$$

Since $S^1$ has nontrivial homology only at dimension $n = 1$. Thus, $H^n(F, M) \cong 0$, as desired $\blacksquare$

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A proof of the fact above can be obtained from lifting the cell structure of $K(G, 1)$ to the universal cover $\widetilde{K(G, 1)}$ and noting that $G$ acts on the $n$-cells on $\widetilde{K(G, 1)}$ freely by deck transformation. Thus, one can use this to turn $C_n(\widetilde{K(G, 1)}; M)$ into $\Bbb Z[G]$-modules.

The cellular boundary maps makes $\{C_\bullet(\widetilde{K(G, 1)}; M)\}$ into a free resolution of $\Bbb Z$ of $\Bbb Z[G]$-modules. Dualizing and taking homology gives $\text{Ext}^\bullet_{\Bbb Z[G]}(\Bbb Z, M)$.

One can modify this resolution by replacing $C_\bullet(\widetilde{K(G, 1)}, M)$ by the subgroup of $G$-invariant elements, so that the dual would have the same homology, but also by definition the homology would be the same as $H^n_{sing}(K(G, 1); M)$. This is roughly how the proof goes.